Problem: Solve for $x$ : $3\sqrt{x} + 4 = 9\sqrt{x} + 3$
Subtract $3\sqrt{x}$ from both sides: $(3\sqrt{x} + 4) - 3\sqrt{x} = (9\sqrt{x} + 3) - 3\sqrt{x}$ $4 = 6\sqrt{x} + 3$ Subtract $3$ from both sides: $4 - 3 = (6\sqrt{x} + 3) - 3$ $1 = 6\sqrt{x}$ Divide both sides by $6$ $\frac{1}{6} = \frac{6\sqrt{x}}{6}$ Simplify. $\dfrac{1}{6} = \sqrt{x}$ Square both sides. $\dfrac{1}{6} \cdot \dfrac{1}{6} = \sqrt{x} \cdot \sqrt{x}$ $x = \dfrac{1}{36}$